Throughout this module the following rules of differentiation will be assumed. (In the table of derivatives below, a is an arbitrary, non-zero constant.) [what is n? integer?]

y |
ax^{n}_{ } |
sin (ax) |
cos (ax) |
e^{ax}_{ } |
ln (ax) |
---|---|---|---|---|---|

dydx |
nax^{n − 1}_{ } |
a cos (ax) |
−a sin (ax) |
a e^{ax}_{ } |
1x |

If a is any constant and u, v are two functions of x, then:

ddx(u + v) |
= | dudx + dvdx |
---|---|---|

ddx(au) |
= | a dudx |

The **stationary points** of a function are those points where the gradient of the tangent (the derivative of the function)
is zero.

**Example 1:** Find the stationary points of the functions:

(a)ƒ(*x*) = 3*x*^{2} + 2*x* − 9

**Solution:** The derivative is: *dy**dx* = 3 × 2*x*^{2−1} + 2 = 6*x* + 2

The stationary points are found by solving the equation:

In this case there is only one solution, *x* = −3. Substituting this into
the equation for ƒ(*x*), the corresponding y value is then:

y |
= | 3(−3)^{2} + 2(−3) − 9_{ } |
---|---|---|

= | 12 |

The stationary point is thus the point with coordinates (−3, 12).

(b)ƒ(*x*)
= *x*^{3} − 6*x*^{2} + 9*x* − 2

**Solution:** The derivative is: *dy**dx* = 3*x*^{2} − 12*x* + 9

The stationary points are found by solving: 3*x*^{2} - 12*x* + 9 = 0.
This is a quadratic equation (see the package on quadratic equations) and may be solved
by factorising:

0 | = | 3x^{2} − 12x + 9 |
= | 3(x^{2} − 4x + 3) |

= | 3(x − 3) (x − 1) |

There are two solutions in this case, *x* = 3 and *x* = 1.
If these are substituted into the function, the two stationary points will be found to be (3, −2) and (1, 2).

**Exercise 1:** Find the stationary points of the following functions:

(a)*y* = 16 − 6*u* − *u*^{2}

**Solution:** The derivative is: *dy**du* = −6 − 2*u*.

The stationary points are found by solving the linear equation:

−6 − 2*u* = 0

In this case there is only one solution *u* = −6 ⁄ 2 = −3.
Substituting this into the expression for the function ƒ(*u*)
= 16 − 6*u* − *u*^{2}, we find the corresponding y value:

The stationary point is thus the point with coordinates (−3, 25).

(b)*y* = 3*x*^{2} − 4*x* + 7

**Solution:**The derivative is: *dy**dx* = 6*x* − 4.

The stationary points are found by solving the linear equation:

6*x* − 4 = 0

In this case there is only one solution *x* = 4 ⁄ 6 = 2 ⁄ 3. Substituting this
this into the expression
for the function ƒ(*x*) = 3*x*^{2} − 4*x* + 7,
we find the corresponding y value:

*y* = 3 × (2 ⁄ 3)^{2} − 4 × 2 ⁄ 3 + 7
= 4 ⁄ 3 - 8 ⁄ 3 + 7 = 17 ⁄ 3

Therefore the function *y* = 3*x*^{2} − 4*x* + 7 has only one
stationary point (2 ⁄ 3, 17 ⁄ 3).

(c)*y* = 13*x*^{3} − *x*^{2} − 3*x* + 2

**Solution:**The derivative is: *dy**dx* = *x*^{2} − 2*x* − 3.

The stationary points are found by solving the quadratic equation *x*^{2} − 2*x* − 3 = 0. Factorisation of this equation gives:

0 = *x*^{2} − 2*x* − 3 = (*x* + 1) (*x* − 3).

This yields two solutions, *x* = −1 and *x* = 3.

Substituting *x* = −1 into the function ƒ(*x*),
the corresponding y value is:

Similarly, substituting *x* = 3:

*y* = 1 ⁄ 3 (3)^{3} − (3)^{2} − 3(3) + 2
= 9 − 9 + 9 + 2 = −7

The two stationary points are thus (−1, 11 ⁄ 3) and (3, −7).

(d)*y* = *x*^{3} − 6*x*^{2} −15*x* + 16

**Solution:**The derivative is: *dy**dx* = 3 *x*^{2} − 12*x* − 15.

The stationary points are found by solving the quadratic equation 3*x*^{2} − 12*x* − 15 = 0. Factorising it:

0 = 3*x*^{2} − 12*x* − 15 = 3(*x*^{2} −4*x* − 5)
= 3(*x* + 1) (*x* − 5)

We obtain two solutions, *x* = *minus;1 and *x* = 5.

Substituting *x* = −1 into the function ƒ(*x*),
the corresponding y value is:

Similarly, substituting *x* = 5:

*y* = (5)^{3} − 6 × (5)^{2} − 15 × (5) + 16
= −1 − 6 + 15 + 16 = −84

The two stationary points are thus (−1, 24) and (5, −84).

Click on questions to reveal solutions

In the package on Introductory Differentiation the height, *h*(*t*), in metres, of a ball thrown vertically at
10m s^{−1}, was given by:

The velocity of the ball, vm s^{−1},
after t seconds, was given by:

The rate of change of velocity with time, which is the **acceleration**, is then given by *a*(*t*), where:

Since the acceleration was derived from *h*(*t*) by two successive differentiations,
the resulting function, which in this case is *a*(*t*),
is called the **second derivative of h(t)** with respect to t.
In symbols:

**Example 2:** Find the first, second and third derivatives of the function: *y* = 3*x*^{6} − 2*x* + 1:

**Solution:**

The **first** derivative is given by:

The **second** derivative is given by:

The **third** derivative is given by:

**Exercise 2:** Find the first, second and third derivatives of the following functions:

(a)*y* = *x*^{3} + e^{2x}

**Solution:**

The **first** derivative is given by:

The **second** derivative is given by:

The **third** derivative is given by:

(b)*y* = sin (*x*)

**Solution:**

The **first** derivative is given by:

The **second** derivative is given by:

The **third** derivative is given by:

Click on questions to reveal solutions

The figure on the right shows part of a function
The point |

Further observations on the gradients of tangents to the curve are:
to the
between
to the
About the |

The rate of change of a function is measured by its derivative. When the derivative is **positive**,
the function is **increasing**, when the derivative is **negative**, the function is **decreasing**.
Thus the rate of change of the gradient is measured by its derivative, which is the **second derivative** of the
original function. In mathematical notation this is as follows:

At the point (a, b):

If | dydx |
= | 0 | If | dydx |
= | 0 |
---|---|---|---|---|---|---|---|

and | d^{2}ydx^{2} |
< | 0 | and | d^{2}ydx^{2} |
< | 0 |

then the point (a, b) is a | then the point (a, b) is a | ||||||

local maximum |
local minimum |

**Example 3: **Find the stationary point of the function *y* = *x*^{2} − 2*x* + 3 and hence determine the nature of this point.

**Solution:**

If
Now
Since
The figure on the right shows the function |

**Exercise 3:** Determine the nature of the stationary points of the following functions:

(a)*y* = 16 − 6*u* − *u*^{2} = 0

**Solution:** From the given function we have:

There is one stationary point when *u* = −3, which is obtained from the equation:

At this point *y* = 16 − 6 × (−3) − (−3)^{2} = 25.
Since *d*^{2}*y**du*^{2} = −2 < 0 for all values of u, this stationary point is a local maximum.
Thus the function *y* = 16 − 6*u* − *u*^{2} has a local
maximum at the point (−3, 25).

(b)*y* = 3*x*^{2} − 4*x* + 7

**Solution:** From the given function we have:

There is one stationary point when *x* = 2 ⁄ 3, which is the solution to the equation:

At this point *y* = 3 × (2 ⁄ 3)^{2} − 4 × (2 ⁄ 3) + 7 = 17 ⁄ 3.
Since *d*^{2}*y**dx*^{2} = 6 > 0 for all values of x, this stationary point is a local minimum.
Thus the function *y* = 3*x*^{2} 4*x* + 7 has a local minimum at the point
(2 ⁄ 3, 17 ⁄ 3).

Click on questions to reveal solutions

**Quiz 1:** Which of the following points is a local maximum of the function y = 2*x*^{3} − 15*x*^{2} − 36*x* + 6?

(a)(−3, 15) Wrong - please try again!

(b)(−6, 25) Wrong - please try again!

(c)(−2, 25) Wrong - please try again!

(d)(−1, 25)Correct - well done!

**Explanation:**

There are two stationary points *x* = −1 and *x* = 6 which are obtained from solving the equation:

0 = *dy**dx* = 6*x*^{2} − 30*x* − 36 = 6 (*x*^{2} − 5*x* − 6 )
= 6 (*x* + 1) (*x* − 6)

At the point *x* = −1 the function value is *y* = 25 and its second
derivative is:

Therefore, the point (−1, 25) is a local maximum.

For the second stationary point, at *x* = 6, the second derivative is:

and *y* = −318. The point (6,−318) is therefore a local minimum.

**Example 4: **Find the stationary point of the function *y* = 2*x*^{3} − 9*x*^{2} + 12*x* − 3 and determine their nature.

**Solution:** From the given function we have:

The stationary points are found by solving:

Now 6*x*^{2} − 18*x* + 12 = 6 (*x*^{2} − 3*x* + 2),
so the solution is found by solving x^{2} − 3*x* + 2 = 0, i.e. x = 1 and *x* = 2 (see the module on Quadratics).
Note that when *x* = 1, *y* = 2 and when *x* = 2, *y* = 1.

When
When The local maximum point at (1, 2) and the local minimum point at (2, 1) are clearly visible in the graph of the given function. |

**Exercise 4:** Determine the nature of the stationary points of the following functions:

(a)*y* = 16 − 6*u* − *u*^{2} = 0

**Solution:** From the given function we have:

There is one stationary point when *u* = −3 which is obtained from the equation:

At this point *y* = 16 − 6 × (−3) − (−3)^{2} = 25.
Since *d*^{2}*y**du*^{2} = −2 < 0 for all values of u, this stationary point is a local maximum.
Thus the function *y* = 16 − 6*u* − *u*^{2} has a local
maximum at the point (−3, 25).

(b)*y* = 13*s*^{2} −12*s* + 32

**Solution:** We have:

There is only one stationary point, when *s* = 18, which is obtained from the equation:

When *s* = 18 the function ƒ(*s*) takes the value *y* = −76. Since its second derivative *d*^{2}*y**ds*^{2} = 23 is positive for all values of s, the stationary point (18, −76) is a **local minimum**.

(c)*y* = *x*^{3} − 6*x*^{2} − 15*x* + 16

**Solution:** We have:

There are two stationary points *x* = −1 and *x* = 5, which
are obtained from solving the quadratic equation:

When *x* = −1 the function value is *y* = 24 and its second derivative *d*^{2}*y**dx*^{2} = 6 × (−1) − 12 = −18 < 0. The point (−1, 24) is a **local maximum**.

At the second point, *x* = 5, the function value is *y* = −84 and the
second derivative *d*^{2}*y**dx*^{2} = 6 × (5) − 12 = 18 > 0. The point (5, −84) is a **local minimum**.

(d)*y* = *x*^{3} − 12*x* + 12

**Solution:** We have:

There are two stationary points *x* = −2 and *x* = 2, which
are obtained from solving the equation:

When *x* = −2 the function value is *y* = 28 and its second derivative *d*^{2}*y**dx*^{2} = 6 × (−2) = −12 < 0. The point (−2, 28) is a **local maximum**.

At the second point, *x* = 2, the function value is *y* = −4 and the
second derivative *d*^{2}*y**dx*^{2} = 6 × (2) = 12 > 0. The point (2, −4) is a **local minimum**.

**Quiz 2:** Given the function *y* = *x* + 1 + 1 ⁄ *x*, which of the following statements is true?

(a)(1, 3) is a local maximum Wrong - please try again!

(b)(1, −3) is a local minimum Wrong - please try again!

(c)(−1, 1) is a local minimum Wrong - please try again!

(d)(−1, −1) is a local maximum Correct - well done!

**Explanation:**

We have: *dy* *dx* = 1 − *x*^{−2} = 1 − 1*x*^{2} and *d*^{2}*y* *dx*^{2} = −(−2)*x*^{−3} = 2*x*^{3}

The stationary points are found from the equation:

and this is zero when 1 − *x*^{2} = 0.
There are two solutions in this case, *x* = −1 and *x* = 1.

When *x* = 1 the function value is *y* = 1 and its second
derivative is *d*^{2}*y**dx*^{2} = 2(−1)^{3} = −2 < 0. The point (−1, −1) is a local maximum.

When *x* = 1 the function value is *y* = 3 and the second derivative is *d*^{2}*y**dx*^{2} = 2(1)^{3} = 2 > 0. The point (1, 3) is a local minimum.

**Quiz 3:** Given the function *y* = *x*^{2}2 − cos (*x*), which of the following statements is true?

(a)(^{π}⁄_{2}, ^{π2}⁄_{8}) is a local maximum Wrong - please try again!

(b)(^{π}⁄_{2}, ^{π2}⁄_{8}) is a local maximum Wrong - please try again!

(c)(0, −1) is a local minimum Correct - well done!

(d)(0, −1) is a local maximum Wrong - please try again!

**Explanation:**

We have: *dy**dx* = *x* + sin (*x*) and *d*^{2}*y**dx*^{2} = 1 + cos (*x*)

The stationary points are found from the equation:

Since sin (0) = 0, a solution to this equation is *x* = 0. At this point the function value is:

Its second derivative when *x* = 0 is:

The point (0, −1) is therefore a **local minimum** of the given function.

In each of the following cases, choose the correct option:

1. If *y* = 3*x*^{2} + 6*x* + 2_{ } then:

(a)(1, 11) is a local minimum

(b) (−1, −1) is a local minimum

(c)(1, 11) is a local maximum

(d)(−1, −1) is a local maximum.

2. The function *x*^{3}3 + *x*^{2}2 − 2 *x* + 4 has a:

(a)local minimum when *x* = 2

(b)local maximum when *x* = −2

(c)local maximum when *x* = 1

(d)local minimum when *x* = −1

3. The function *y* = sin (*x*) + cos (*x*) has a:

(a)local maximum when *x* = ^{π}⁄_{4}

(b)local minimum when *x* = ^{π}⁄_{4}

(c)local maximum when *x* = ^{π}⁄_{2}

(d)local minimum when *x* = ^{π}⁄_{2}